Question: Let $y=\sqrt{e^x}$. Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\sqrt{e^x}}{2}$ (Choice B) B $x\sqrt{e^{x-1}}$ (Choice C) C $\dfrac{1}{2\sqrt{e^x}}$ (Choice D) D ${\sqrt {e^x}}$
Explanation: $\sqrt{e^x}$ is a composition of two, more basic, functions: $e^x$ and $\sqrt x$. In other words, suppose $u(x)=e^x$ and $v(x)=\sqrt x$, then $\sqrt{e^x}=v\bigl(u(x)\bigr)$, or $(v\circ u)(x)$. Therefore, $\dfrac{dy}{dx}$ can be found using the chain rule : $\begin{aligned} \dfrac{d}{dx}\left[v\Bigl(u(x)\Bigr)\right]&=\dfrac{dv}{du}\cdot\dfrac{du}{dx} \\\\ &=v'\Bigl(u(x)\Bigr)\cdot u'(x) \end{aligned}$ Finding $v'\Bigl(u(x)\Bigr)$ $v(x)=\sqrt x$, and therefore $v'(x)=\dfrac{1}{2\sqrt x}$. Now we plug $u(x)=e^x$ into $v'$ : $\begin{aligned} v'\Bigl(u(x)\Bigr)&=v'\Bigl(e^x\Bigr) \\\\ &={\dfrac{1}{2\sqrt{e^x}}} \end{aligned}$ Finding $u'(x)$ $u(x)=e^x$, and therefore $u'(x)={e^x}$. Putting things together $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}\left(\sqrt{e^x}\right) \\\\ &=\dfrac{d}{dx}\left[v\Bigl(u(x)\Bigr)\right]&&\gray{\text{Let }u(x)=e^x\text{, }v(x)=\sqrt x} \\\\ &=v'\Bigl(u(x)\Bigr)\cdot u'(x)&&\gray{\text{The chain rule}} \\\\ &={\dfrac{1}{2\sqrt{e^x}}}\cdot {e^x} \\\\ &=\dfrac{\sqrt{e^x}}{2} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\dfrac{\sqrt{e^x}}{2}$.